【题目】如图,菱形ABCD中,∠BAD=120°,点M为BC上一点,点N为CD上一点,若△AMN有一角等于60°,试判断△AMN的形状,并说明理由DNBMC
23.如图,已知正方形ABCD中,点M、N分别在边BC、CD上,∠MAN=45°DNBMc(1)求证:MN=BM+DN(2)当AB=6,MN=5时,求△CMN的面积
Ideally, dN/dS should be estimated at every site to find evidence of selection (which is only possible when comparing more than two species in a phylogenetic context) and not averaged over the entire gene. However, an over-representation of non-synonymous substitutions can be used as a crude...
型号 BMC/135KB BMC30/135KB BMC30K DN-SCANNER ———深圳长欣自动化设备有限公司——— *国外直接采购,正品行货,价格优惠,售后保修,型号齐全,优势库存。 *所有备件的质保期均为1年,经过专业测试认证。 *如果您需要订购产品超过1件,请与我们联系,我们可以为您提供折扣。 *我们仅采用顺丰快递方式递送备件。
et al. Efficacy and safety of Zolbetuximab plus chemotherapy for advanced CLDN18.2-positive gastric or gastro-oesophageal adenocarcinoma: a meta-analysis of randomized clinical trials. BMC Cancer 24, 240 (2024). https://doi.org/10.1186/s12885-024-11980-w Download citation Received02 November ...
如图.AC平分∠BAD.CM⊥AB于点M.CN⊥AD交AD延长线于点N.若BM=DN.那么∠ADC与∠ABC的关系是( )A. 相等 B. 互补C. 和为150° D. 和为165° B [解析]∵AC平分∠BAD.CM⊥AB于点M.CN⊥AN. ∴CM=CN.∠CND=∠BMC=90°. ∵BM=DN. 在△CND与△CMB中. ∵ . ∴△CND≌△CMB.
【题文】设四边形ABCD 为平行四边形, | (AB)|=6,| (AD)|=4 .若点M,N 满足 (BMC)=3 (DN)=2 (NC) ,则 AM ⋅ (NM) =
13[分析]过点C作CN⊥AD交AD延长线于点N由角平分线的性质得到CN=CM然后证明△CDN≌△CBM得到DN=BMCD=CB=25然后求出AN=AM=4则AD=4DN即可求出四边形的周长[详解]解析:13[分析]过点C作CN⊥AD,交AD延长线于点N,由角平分线的性质,得到CN=CM,然后证明△CDN≌△CBM,得到DN=BM,CD=CB=2.5,然后求出AN=...
阿里巴巴为您找到水防水锤止回阀DN100 球形止回阀HQ44X-16C无磨损球式止回阀生产商,提供相关止回阀工厂资质及电话联系,产品图片,价格/报价,品牌/贴牌等企业信息,卖家可在线找采购订单,买家可找工厂沟通批发定制需求,一站式工厂寻源、厂家直销平台
The investigational use of zolbetuximab (IMAB362), a groundbreaking monoclonal antibody medication targeting claudin 18.2 (CLDN18.2), for treatment of advanced gastrointestinal cancers is currently underway. The unclear clinicopathological characteristics and tumour immune microenvironment of CLDN18.2-positive...